Question:

Step 1: To solve the given problem, we first need to understand the properties and behavior of the function $f(x)$ as defined. The function $f$ is differentiable and defined on the interval $\left[0,\dfrac{\pi}{2}\right]$ with $f(x) > 0$ for all $x$ in this interval. Step 2: We are given the equation $f(x)+\displaystyle\int_0^x f(t)\sqrt{1-(\log_e f(t))^2}\,dt=e$. To proceed, let's differentiate both sides of this equation with respect to $x$ to simplify and understand the relationship between $f(x)$ and its derivative. Step 3: Differentiating the given equation yields: $$\frac{d}{dx}\left(f(x)+\displaystyle\int_0^x f(t)\sqrt{1-(\log_e f(t))^2}\,dt\right) = \frac{d}{dx}(e)$$ Using the fundamental theorem of calculus, we get: $$f'(x) + f(x)\sqrt{1-(\log_e f(x))^2} = 0$$ Step 4: To simplify the equation from Step 3 and relate it to the given options, let's rearrange and solve for $f'(x)$: $$f'(x) = -f(x)\sqrt{1-(\log_e f(x))^2}$$ This equation shows the relationship between the derivative of $f(x)$ and $f(x)$ itself, involving the logarithmic term. Step 5: Now, we need to find a way to express $f(x)$ in terms of $x$ or directly evaluate $\left(6\log_e f\!\left(\dfrac{\pi}{6}\right)\right)^2$. Given the complexity of directly solving for $f(x)$, let's consider the properties of the logarithmic function and its relation to the given equation. Step 6: Let $g(x) = \log_e f(x)$. Then, $f(x) = e^{g(x)}$. Substituting into the equation from Step 4 gives: $$e^{g(x)}g'(x) = -e^{g(x)}\sqrt{1-(g(x))^2}$$ Simplifying, we find: $$g'(x) = -\sqrt{1-(g(x))^2}$$ Step 7: The equation $g'(x) = -\sqrt{1-(g(x))^2}$ can be solved by separating variables. However, recognizing that this is a separable differential equation, let's solve it directly: $$\frac{dg}{dx} = -\sqrt{1-g^2}$$ Separating variables gives: $$\frac{dg}{\sqrt{1-g^2}} = -dx$$ Integrating both sides yields: $$\arcsin(g) = -x + C$$ where $C$ is the constant of integration. Step 8: To find $g(x)$, we solve the equation from Step 7 for $g$: $$g(x) = \sin(-x + C)$$ Since $g(x) = \log_e f(x)$, we have: $$\log_e f(x) = \sin(-x + C)$$ Given $f(0) > 0$ and the initial condition from the problem, we can determine $C$. Step 9: Using the initial condition $f(0) + \displaystyle\int_0^0 f(t)\sqrt{1-(\log_e f(t))^2}\,dt = e$, which simplifies to $f(0) = e$, we find $\log_e f(0) = 1$. Thus, $\sin(C) = 1$, implying $C = \frac{\pi}{2}$. Step 10: Substituting $C = \frac{\pi}{2}$ into the equation for $\log_e f(x)$ gives: $$\log_e f(x) = \sin\left(-x + \frac{\pi}{2}\right) = \cos(x)$$ Therefore, $\log_e f\!\left(\dfrac{\pi}{6}\right) = \cos\!\left(\dfrac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. Step 11: Finally, to find $\left(6\log_e f\!\left(\dfrac{\pi}{6}\right)\right)^2$, we substitute the value of $\log_e f\!\left(\dfrac{\pi}{6}\right)$: $$\left(6\log_e f\!\left(\dfrac{\pi}{6}\right)\right)^2 = \left(6 \cdot \frac{\sqrt{3}}{2}\right)^2 = (3\sqrt{3})^2 = 27$$ Thus, $\left(6\log_e f\!\left(\dfrac{\pi}{6}\right)\right)^2$ is equal to 27.